



 
Hi Thys, I found an expression for what you want on the Mechanical Engineering web site at http://www.aboutmech.com/2014/06/equationofpathofprojectile.html. It is item number 5. on their page. I don't think however it is in a form that will be particularly useful to you. I redrew the diagram on their site and labeled it to be more consistent with the notation you are using The cannon is at $O,$ the target is at $B,$ $d$ is the distance to the target along the line of sight $OB$ and $h$ is the height of the target above the horizontal. The distances are in feet and the angles are measured in degrees. The line segment $OA$ is in the direction of the initial velocity vector of the cannon ball, at an angle of $\alpha$ degrees to the horizontal. With this notation the formula given on the Mechanical Engineering web site is \[d = \frac{u^2}{G \; \cos^2 \beta} \left[ \sin[2(\alpha  \beta)  \beta]  \sin \beta\right] = \frac{u^2}{G \; \cos^2 \beta} \left[ \sin[2 \alpha  3 \beta)]  \sin \beta\right]\] where $G$ is the force due to gravity ad $u$ is the initial velocity. Using the facts that \[\sin(2 \alpha  3 \beta) = \sin 2 \alpha \cos 3 \beta + \cos 2 \alpha \sin 3 beta\] \[\sin 3 \beta = 3 \sin \beta  4 \sin^3 \beta\] \[\cos 3 \beta = 4 \cos^3 \beta  3 \cos \beta\] and reading the values of $\sin \beta$ and $\cos \beta$ from the diagram the expression for $d$ above becomes \[d = \frac{u^2}{G \frac{R^2}{d^2}} \left[ \left( 4 \frac{R^3}{d^3}  3 \frac{R}{d} \right) \sin 2 \alpha + \left(3 \frac{h}{d}  4 \frac{h^3}{d^3}\right) \cos 2 \alpha  \frac{h}{d}\right].\] Since $R^2 + h^2 = d^2$ this simplifies to \[1 = \frac{u^2}{G R^2} \left[ \frac{R^3  3 R h^2}{R^2 + h^2} \sin 2 \alpha + \frac{3 h R^2  h^2}{R^2 + h^2} \cos 2 \alpha  h \right] \] or \[u = \sqrt{ \frac{G R^2}{\left[ \frac{R^3  3 R h^2}{R^2 + h^2} \sin 2 \alpha + \frac{3 h R^2  h^2}{R^2 + h^2} \cos 2 \alpha  h \right] }} \] This expression looks onerous but when $h = 0,$ the cannon and the target are at the same altitude it reduces to the expression you sent \[u = \sqrt{\frac{RG}{\sin 2 \alpha}}.\] For the example you sent with $\alpha = 45^o$ you have $\sin 2 \alpha = \sin 90^o = 1,$ and $\cos \alpha = \cos 90^o$ so \[u = \sqrt{ \frac{GR^2 (R^2 + h^2)}{R^3  3 R h^2}}\] I hope this helps,  


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